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Question:(II) (a) How much energy is transformed by a typical 65-kg person who runs at\({\bf{15 km/h}}\)for\({\bf{30 min/day}}\)in one week (Table 15–2)? (b) How many food calories would the person have to eat to make up for this energy loss?

Short Answer

Expert verified
  1. The energy transformed by the person in one week is\(1.45 \times {10^7}{\rm{ J}}\).
  2. The energy utilized to make up for this energy loss is 3460.62 Cal.

Step by step solution

01

Identification of given data

The given data from table 15-2 can be listed below as:

  • The metabolic rate of the person while running 15 km/h is

\(E = 1150{\rm{ W}}\left( {\frac{{1{\rm{ J/s}}}}{{1{\rm{ W}}}}} \right) = 1150{\rm{ J/s}}\).

  • The time duration is 30 min/day a week.
02

Understanding the transformation of the energy of a person while running

The person is running; so he/she requires mechanical energy. The person utilizes the chemical energy of the food.

This chemical energy transforms into mechanical energy.

For maintaining the person's temperature, this chemical energy is also being converted into thermal energy.

03

(a) Determination of the energy transformed by the person in a week

One week has seven days. The person is runs once every day. He/she takes30 minutes for one run. So, he/she is running seven times a week.

The energy transformed by the person in a week can be expressed as:

\(\begin{aligned}{c}E &= 1150{\rm{ J/s}}\left( {\frac{{60{\rm{ s}}}}{{1{\rm{ min}}}}} \right)\left( {\frac{{30{\rm{ min}}}}{{{\rm{ 1 run}}}}} \right)\left( {\frac{{7{\rm{ times}}}}{{1{\rm{ week}}}}} \right)\\ &= 1.45 \times {10^7}{\rm{ J/week}}\end{aligned}\)

Thus, the energy transformed by the person in one week is \(1.45 \times {10^7}{\rm{ J}}\).

04

(b) Determination of the calories utilized by the person to make this energy loss

The food energy utilized by the person can be expressed as:

\(\begin{aligned}{c}E &= 1.45 \times {10^7}{\rm{ J}}\left( {\frac{{1{\rm{ Cal}}}}{{4.19 \times {{10}^3}{\rm{ J}}}}} \right)\\ &= 3460.62{\rm{ Cal}}\end{aligned}\)

Thus, the energy utilized to make up for this energy loss is 3460.62 Cal.

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